Question

How do you find the limit of $\dfrac{x}{{x - 1}} - \dfrac{1}{{\ln (x)}}$ as $x$ approaches $1$ ?

Answer 1

In order to solve this problem, you must know if you've got an indeterminate form for your answer to your limit, then you'll take the derivative of the numerator and of the denominator separately so as to search out the limit. You can repeat this process if you still get an indeterminate form. An indeterminate form is when the limit seems to approach a deeply weird answer.

Complete step by step solution:
We have given,
$\dfrac{x}{{x - 1}} - \dfrac{1}{{\ln (x)}}$
We have to find the limit as $x$ approaches $1$ ,
Therefore, we can write this as,
$\Rightarrow \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{x}{{x - 1}} - \dfrac{1}{{\ln (x)}}} \right) = \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{x(\ln x) - x + 1}}{{(x - 1)(\ln x)}}} \right)$
As the limit $x$ approaches $1$ , we will get,
\dfrac{{1(\ln 1) - 1 + 1}}{{\ln 1 - 1(\ln 1)}}$$$ = \dfrac{0}{0}$ Since we get the indeterminate form that is $\dfrac{0}{0}$ . Now, we have to use L'Hôpital's Rule that is if you've got an indeterminate form for your answer to your limit, then you'll take the derivative of the numerator and of the denominator separately so as to search out the limit. Applying L’Hopital’s Rule , \Rightarrow \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{x(\ln x) - x + 1}}{{(x - 1)(\ln x)}}} \right) = \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{\dfrac{d}{{dx}}\left( {x(\ln x) - x + 1} \right)}}{{\dfrac{d}{{dx}}\left( {x\ln x - \ln x} \right)}}} \right)$$ = \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{\ln x}}{{\ln x + 1 - \dfrac{1}{x}}}} \right) As the limit $x$ approaches $1$ , we will again get the indeterminate form that is $\dfrac{0}{0}$ . Therefore, again applying L’Hopital’s Rule , \Rightarrow \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{\ln x}}{{\ln x + 1 - \dfrac{1}{x}}}} \right) = \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{\dfrac{d}{{dx}}(\ln x)}}{{\dfrac{d}{{dx}}\left( {\ln x + 1 - \dfrac{1}{x}} \right)}}} \right)$$
$= \mathop {\lim }\limits_{x \to 1} \left( {\dfrac{{\dfrac{1}{x}}}{{\dfrac{1}{x} + \dfrac{1}{{{x^2}}}}}} \right) = \dfrac{1}{2}$
As the limit $x$ approaches $1$, we will get $\dfrac{1}{2}$ .
We get the required result.

Note:
If I had found the solution to still in indeterminate form, then I might need to continue using L'Hôpital's Rule. But as soon as I get a determinate form , I have to stop. Because when the solution is not any longer an indeterminate form, L'Hôpital's Rule does not apply.